Q4. For the gas reaction A(g) ⇌ 2B(g), ΔrG⊖ = 0 at 298 K. Is a mixture with pA = pB = 2 bar at equilibrium? If not, which way does it go?
Kp = e0 = 1 (pressures in bar). Q = 22/2 = 2 > K, so ΔrG = RT ln 2 = 2477.6 × 0.693 ≈ +1.72 kJ mol−1: not at equilibrium; it moves backward, B forming A.
Q5. Using Kc = 3.81 × 10−3 for the phosphorylation of glucose at 298 K (NCERT Problem 6.10), find ΔrG for an idealised mixture in which Q = 1.0 × 10−5.
ΔrG = RT ln(Q/K) = 8.314 × 298 × ln(1.0 × 10−5/3.81 × 10−3) = 2477.6 × (−5.943), about −14.7 kJ mol−1. With so little product present, the reaction goes forward even though ΔrG⊖ is positive.
Q6. For an illustrative biochemical calculation, take transformed standard Gibbs energies at the same fixed pH 7 and 298 K: +13.8 kJ mol−1 for uncoupled glucose phosphorylation and −30.5 kJ mol−1 for ATP hydrolysis. Estimate the transformed standard change ΔrG°′ and corresponding K′ for the coupled reaction. The prime denotes the fixed-pH biochemical convention; do not combine values from mismatched standard states.
Adding the two reactions adds their Gibbs energy changes: 13.8 + (−30.5) = −16.7 kJ mol−1. K = e16 700/2477.6 = e6.74, about 8.5 × 102, so the coupled reaction favours the products. (Equivalently, the two constants multiply.)
Q7. For N2(g) + 3H2(g) ⇌ 2NH3(g), ΔrH⊖ = −92.2 kJ mol−1 and ΔrS⊖ = −198.7 J K−1 mol−1, taken as constant. At about what temperature is K = 1?
K = 1 when ΔrG⊖ = ΔrH⊖ − TΔrS⊖ = 0, so T = 92 200/198.7, about 464 K. Below this K > 1; above it K < 1.
Q8. Sucrose hydrolysis has ΔrG⊖ ≈ −76 kJ mol−1 (NCERT Problem 6.11), yet uncatalysed sucrose hydrolysis can be very slow in a sterile near-neutral solution. Explain, and say what happens if the enzyme invertase is added.
ΔrG⊖ fixes where equilibrium lies (K ≈ 2 × 1013, almost complete hydrolysis), not how fast it is reached. Without a catalyst the reaction is extremely slow. Invertase speeds it up, so hydrolysis becomes faster under conditions where the enzyme is active; the enzyme does not change K.